infimum and supremum for real numbersInfimumAndSupremumForRealNumbers2006-02-19 15:04:472006-04-26 11:04:48Topictopic entry on real numbers% this is the default PlanetMath preamble. as your knowledge
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\newtheorem{thm}{Theorem}
\newtheorem{defn}{Definition}
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\newtheorem{lemma}{Lemma}
\newtheorem{cor}{Corollary}Suppose $A$ is a non-empty subset of $\R$.\, If $A$ is bounded from above, then
the axioms of the real numbers imply that there exists
a \emph{least upper bound} for $A$.
That is, there exists an $m\in \R$ such that
\begin{enumerate}
\item $m$ is an upper bound for $A$, that is, $a\le m$ for all $a\in A$,
\item if $M$ is another upper bound for $A$, then $m\le M$.
\end{enumerate}
Such a number $m$ is called the \emph{supremum} of $A$,
and it is denoted by $\sup A$. It is easy to see that
there can be only one least upper bound. If $m_1$ and $m_2$ are
two least upper bounds for $A$. Then $m_1\le m_2$ and $m_2\le m_1$,
and $m_1=m_2$.
Next, let us consider a set $A$ that is bounded from below. That is, for
some $m\in \R$ we have $m\le a$ for all $a\in A$.\, Then we say that
$M\in \R$ is a
a \emph{greatest lower bound} for $A$ if
\begin{enumerate}
\item $M$ is an lower bound for $A$, that is, $M \le a$ for all $a\in A$,
\item if $m$ is another lower bound for $A$, then $m\le M$.
\end{enumerate}
Such a number $M$ is called the \emph{infimum} of $A$,
and it is denoted by $\inf A$. Just as we proved that the supremum
is unique, one can also show that the infimum is unique.
The next lemma shows that the infimum exists.
\begin{lemma} Every non-empty set bounded from below has a greatest lower bound.
\end{lemma}
\begin{proof} Let $m\in \R$ be a lower bound for non-empty set $A$.
In other words, $m\le a$ for all $a\in A$. Let
$$
-A = \{ -a \in \R : a\in A\}.
$$
Let us recall the following result from \PMlinkname{this page}{InequalityForRealNumbers};
if $m$ is an upper(lower) bound for $A$, then $-m$ is a lower(upper)
bound for $-A$.
Thus $-A$ is bounded from above by $-m$.
%In fact, if $b\in B$, then
%$b=-a$ for some $a\in A$, so $-b=a\ge m$, and $b\le -m$.)
It follows that $-A$ has a least upper bound $\sup (-A)$.
Now $-\sup (-A)$ is a greatest lower bound for $A$.
First, by the result, it is a lower bound for $A$.
Second, if $m$ is a lower bound for $A$,
then $-m$ is a upper bound for $-A$, and
$\sup (-A)\le -m$, or $m\ge -\sup (-A)$.
\end{proof}
The proof shows that if $A$ is non-empty and bounded from below,
then
$$
\inf A = -\sup (-A).
$$
In consequence, if $A$ is bounded from above,
then
$$
\sup A = -\inf (-A).
$$
In many respects, the supremum and infimum are similar to the maximum
and minimum, or the largest and smallest element in a set.
However, it is important to notice that the $\inf A$ and $\sup A$
do not need to belong to $A$. (See examples below.)
\subsubsection*{Examples}
\begin{enumerate}
\item
For example, consider the set of negative real numbers
$$
A = \{ x\in \R:\,\, x<0\}.
$$
Then\, $\sup A = 0$. Indeed. First, $a < 0$ for all $a \in A$,
and if $a < b$ for all $a \in A$, then $0 \le b$.
\item The sequence \,\,
$-(1\!-\!\frac{1}{1}),\,1\!-\!\frac{1}{2},\,-(1\!-\!\frac{1}{3}),\, 1\!-\!\frac{1}{4},\,-(1\!-\!\frac{1}{5}),\,...$\,\,
is not convergent.\, The set\, $A = \{(-1)^n(1-\frac{1}{n}):\,\, n\in\mathbb{Z}_+\}$\, formed by its members has the infimum $-1$ and the supremum 1.
\end{enumerate}